Power Supply Resevoir Size
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gootee
diyAudio Member

Join Date: Nov 2006
Location: Indiana
I think I just did something that could be useful and maybe even interesting. It's about calculating reservoir capacitances.

See: http://www.diyaudio.com/forums/chip-...ml#post3403933

Attached is an image of a spreadsheet that I threw together, to turn the crank on the equations, for me.

It has a plot of the upper and lower bounds of the minimum required reservoir capacitance vs the Max Rated Output Power (to just barely avoid clipping).

given the unloaded rail voltage, the Vclip voltage (across the amp), which is given in the LM3875 datasheet in this case, and Rload, fmains, and an increment to change the resulting Vpeak_max voltage by, and maybe some other things; just take a look.

Basically, I was able to express everything (e.g. peak current and max ripple voltage) in terms of one unknown variable, such as the max peak output voltage, plus a few variables that become fixed parameters for any particular case.

I was actually able to eliminate the need to manually select or calculate or guess at the acceptable ripple voltage amplitude. It's just calculated in terms of Vrail_unloaded, Vclip, and Vpeak_max (output peak voltage), the first two of which are assumed to be known.

However, you can still select the capacitance value based on the p-p ripple, or based on the Vpeak_max output voltage, or based on the Max Rated Output Power. Those are all dependent on only one of the others. i.e. Know one and you know them all. So I just sweep the Vpeak_max and calculate the minimum required C for each step, and then also calc the equivalent ripple amplitude and Max Rated Output Power.

And the fixed parameters in the upper left corner can just be changed at will, and everything automatically recalculates.

I DO HOPE that someone will check the derivation for accuracy, at the link given above! Then maybe I can post a similar spreadsheet.

Cheers,

Tom
Attached Images
 upper_lower_C_exmpl.jpg (315.7 KB, 149 views)

Last edited by gootee; 10th March 2013 at 08:44 AM.

 10th March 2013, 09:58 AM #1772 AndrewT   diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders I can't see the spreadsheet download. __________________ regards Andrew T.
 10th March 2013, 01:09 PM #1773 gootee   diyAudio Member   Join Date: Nov 2006 Location: Indiana I have only posted an image of it, so far.
 10th March 2013, 01:50 PM #1774 PauloPT   diyAudio Member   Join Date: Jan 2010 Location: Lisboa, Portugal Am I reading this right - a 50W amplifier would need 288mF capacitance to sound good? __________________ Kind Regards, Paulo.
 10th March 2013, 01:56 PM #1775 AndrewT   diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders No. The sensible maximum when the open circuit rail voltage is @ +-33Vdc is about 45W. A more practical maximum would be around 40W from +-33Vdc supply rails. We are back to the same old idea of asking a too low voltage transformer to deliver impossibly high claimed power figures. If you want 50W then use a higher rail voltage. If you want 60W then the next step up in rail voltage is warranted. The spreadsheet would make the number crunching very easy. __________________ regards Andrew T.
 10th March 2013, 02:10 PM #1776 PauloPT   diyAudio Member   Join Date: Jan 2010 Location: Lisboa, Portugal Then a 40W amp would need 288mF... that's a lot! __________________ Kind Regards, Paulo.
 10th March 2013, 02:13 PM #1777 AndrewT   diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders Read the graphs !!!! 40W is modeled as needing between 14.5mF and 20mF, not 288mF __________________ regards Andrew T.
jacco vermeulen
diyAudio Member

Join Date: Oct 2004
Location: At the sea front, Rotterdam or Curaçao
Quote:
 Originally Posted by PauloPT that's a lot!
Alas, you do not see how much difference 1.5V can make.
__________________
Why does it hurt when I Pee ? (Ay Ay Ay https://www.youtube.com/watch?v=w5fLGWrHouk)

PauloPT
diyAudio Member

Join Date: Jan 2010
Location: Lisboa, Portugal
Quote:
 Originally Posted by AndrewT Read the graphs !!!! 40W is modeled as needing between 14.5mF and 20mF, not 288mF
You're right. But anyway I was not expecting to see such high values when the power is just a little more.
__________________
Kind Regards,
Paulo.

 10th March 2013, 03:44 PM #1780 gootee   diyAudio Member   Join Date: Nov 2006 Location: Indiana Keep in mind that the resulting capacitances are not what is needed "to sound good". They are the absolute minimum capacitance needed to prevent any clipping at all, when the amp is producing its MAXIMUM rated output power. Actually, they determine the max rated output power. Why are some of them so large? Basically, I just selected some of the Vpeak_max values to show the impractical cases, just to illustrate the concepts. As the Vpeak_max is pushed up, to get a higher max rated output power, there is less voltage space for the ripple to occupy. The maximum rail voltage is what it is, and the amplifier's Vclip figure (between +rail pin and output pin) can't occupy less than 4.5V (when the rail is 33V, according to datasheet) of the voltage space between the signal maximum and the ripple minimum. When you try to increase Vpeak_max too much and start to run out of room in the voltage space, the ripple amplitude needs to get stupidly small in order to fit, and not cause clipping, so the capacitance gets stupidly large. You would probably never want to try to cut it that close, anyway. Note, too, that the upper-bound C value is calculated based on an absolutely-worst-case output current, i.e. the current is DC, but at the maximum sine peak current that the amplifier should be capable of producing. So it's 1.414 times higher than the maximum RMS current that the amp should be expected to produce. I originally settled on that type of DC output (i.e. at the maximum peak sine level) as being the worst case, way back when I was worried about different phase relationships between the charging pulses and the sine signal, since the DC would cover all of those. But in thinking about lower and lower sine signal frequencies, much lower than the charging pulses' repetition rate, with the worst case (integral of the current waveform) being when the sine's peak was centered between charging pulses, it became obvious that as the sine signal frequency went toward zero and its peak grew very wide, its waveform, with peak centered between charging pulses, would approach DC at the sine's peak level. At any rate, it seems intuitively clear that the upper-bound capacitance that is derived in that way will always be able to supply the current needed, for any signal waveform with a voltage amplitude that never exceeds Vpeak_max. BUT, in real music, e.g. live music, with more than one tone or instrument waveform present at the same time, there could often be coincident peaks, where the amplitudes would sum. This analysis had to assume that no voltage peaks exist, in the input waveform, would cause the output to need to be greater than Vpeak_max volts. Last edited by gootee; 10th March 2013 at 04:06 PM.

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